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A ball of mass 10 kg is attached by a lightweight
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Answer #1

Lets

m_1g-T=m_1a

T-m_2gsin\theta=m_2a

Adding both equations we get ,

(m_1-m_2sin\theta)g=(m_1+m_2)a

\implies a=2.13m/s^2

So, a=

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